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Mixture replacement Problems:How to solve effectively and q

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Mixture replacement Problems:How to solve effectively and q

by prab.sahi06 » Mon Jul 08, 2019 9:38 am
A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?

a. 2:3
b. 1:2
c. 1:3
d. 3:4
e. 1:1

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by oabpyhfv » Sat Jul 20, 2019 12:22 am
Since you know AG = 3CE and DG = 3DE ...eq(1),

Now the area of the top most triangle(CDE)
=1/2(EC x DE) = 42 ...eq(2) (given)

And Now, consider the area of the entire triangle(ADG) = 1/2(AG x DG)
= 1/2(3EC x 3DE)-------- replace the values using (eq(1))
= 9/2(EC x DE)
= 9(1/2(EC x DE)) ---eq(3)--- in this 1/2(EC x DE) is equal to 42(given) i.e. eq(2) replace it in the eq(3) with the value 42.
= 9 (42) = 378.

I hope this njmcdirect is helpful.

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by shrey.me[email protected] » Fri Oct 25, 2019 5:19 am
prab.sahi06 wrote:A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?

a. 2:3
b. 1:2
c. 1:3
d. 3:4
e. 1:1

At first, the total mixture of 100 litres (80+20)
the proportion of water in the mixture is 1/5 and milk is 4/5

then, he sold 1/4th of 100 litres i.e 25 litres of the mix.
Now, remaining 75 litres, has 1/5th water, too
equals 15 litres of water, and
4/5th of milk = 60 litres of milk.

He adds another 25 litres of water, to make for the sold mix...
therefore, 25+15 = 40 litres of water now and milk, as you know (unchanged) is 60 litres...
40/60 = 2/3 or 2:3

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by nitink » Tue Nov 19, 2019 10:14 am
Total mixture = 100 litres (80+20)
the proportion of water in the mixture is 20/100 = 1/5 and milk is 80/100 = 4/5

He sold 1/4th of 100 litres i.e 25 litres of the mixture.
Now, remaining 75 litres, has 1/5th water.

1/5*75 = 15 litres of water

milk = 75 -15 = 60 litres

New mixture has 25+15 = 40 litres of water now and milk remains unchanged at 60 litres.
so, ratio = 40/60 = 2/3

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by [email protected] » Tue Nov 19, 2019 6:56 pm
Hi All,

We're told that a milkman mixes 20 liters of water with 80 liters of milk and that after selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. We're asked for the ending proportion of water to milk. The math behind this question can be done in a number of different ways, but you might find it easiest to work in small calculations as you work through it.

To start, we have 100 total liters; the ratio of water to milk in that mix is 20:80 = 1:4. This means that for every 1 liter of water, there are 4 liters of milk.

1/4 of the mixture = (1/4)(100) = 25 liters... so these 25 liters consist of 5 liters of water and 20 liters of milk. After that portion is sold, we're left with a mixture that is 15 liters of water and 60 liters of milk. Adding 25 liters of water to this mixture gives us....

40 liters of water and 60 liters of milk.... 40:60 = 4:6 = 2:3, so the final mixture of water to milk is 2:3.

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Rich
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by SampathKp » Tue Dec 17, 2019 8:06 pm
prab.sahi06 wrote:A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?

a. 2:3
b. 1:2
c. 1:3
d. 3:4
e. 1:1
M . W . T
80 . 20 . 100 . The mixture has water is to milk in the ratio .1:4
75 . Selling 1/4 of the mixture means milk man has sold 25 ltrs (100/4)
60 . 15 . 75 . After selling 25 ltrs, the 75 ltrs of remaining mixtures has 60 ltrs of milk & 15 ltrs of water. Same proportion of water is ti milk of 1:4 (i.e water is 1/5 x75 =15. Milk is 60 (75-15)
100 , Now to make the mixture to 100 Ltrs, milkman added 25 ltrs of water
60 . 40 . 100 So new 100 ltrs of mixture has 60 Ltrs of milk and 40 ltrs of water .

Current proportion of water to milk is 40: 60 i.e 2:3

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Re: Mixture replacement Problems:How to solve effectively and q

by gmatknight » Thu Feb 27, 2020 10:29 pm
People, do not fall in the trap of forcing variables into formulas. It's annoying, I know, but understand the concept at play. This is not a difficult question in my view, but one could make it terribly difficult for no reason.

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Re: Mixture replacement Problems:How to solve effectively and q

by PhillipZhang87 » Tue Sep 15, 2020 3:41 pm
Water: Milk: Total
20: 80: 100 Original Porportion
15: 60: 75 3/4 * Original Porportion
40 : 60: 100 New porportion
2: 3 : 5

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